# Finding energy eigenvalue of a harmonic oscillator using a Hamiltonian

1. Oct 10, 2012

### JordanGo

1. The problem statement, all variables and given/known data
Find the energy eigenvalue.

2. Relevant equations

H = (p^2)/2m + 1/2m(w^2)(x^2) + λ(x^2)

Hψ=Eψ

3. The attempt at a solution

So this is what I got so far:

((-h/2m)(∂^2/∂x^2)+(m(w^2)/2 - λ)(x^2))ψ=Eψ

I'm not sure if I should solve this using a differential equation method, or is there an easier trick?

Thank you!

2. Oct 11, 2012

### Simon Bridge

... you should solve using the DE methods.

Since the potential is in x only, the partials become exact[1]:
$$\left ( \frac{-\hbar^2}{2m}\frac{d^2}{dx^2} + \zeta x^2 - E \right )\psi(x) = 0 : 2\zeta = m\omega^2+2\lambda$$... and it is really useful for these things to learn to use $\LaTeX$ :)

I don't know why you didn't square your $\hbar$ or why you have a $-\lambda$ in there but the setup has a $+\lambda$.
Also, I'm guessing by "$h$" you meant $\hbar=h/2\pi$ since the momentum operator is $-i\hbar\frac{\partial}{\partial x}$.

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[1] if you are expected to do this in 3D, then you'll have to deal with the other two directions.

Last edited: Oct 11, 2012
3. Oct 11, 2012

### JordanGo

Ok, so now I tried to perform the differential equation, but then I get an equation in the form of:

Ψ=Ae(ς-E)x+Be(-(ς-E)x)

How can I find the energy eigenvalue from this equation?

4. Oct 11, 2012

### Simon Bridge

You put it into the Schrodinger equation and solve for E.

But I am not sure that wavefunction is a solution... don't you associate Gaussians with harmonic oscillator potentials?

5. Oct 11, 2012

### JordanGo

We haven't done any sort of Gaussian...

But I came up with another way to solve, can you see if this makes sense:

I can write the Hamiltonian as:

$H =\frac{p^2}{2m} + x^2 (\frac{m\varpi^2}{2} + \frac{\lambda}{\sqrt{2}})$

or

$H = \frac{p^2}{2m} + \frac{mω'^2x^2}{2}$

where

$ω'=\sqrt{ω^2+\frac{\sqrt{2}λ}{m}}$

then I can use

$E = (\hbar)ω'(n+\frac{1}{2})$

6. Oct 11, 2012

### Simon Bridge

Sure - you can exploit existing solutions. What what stopping you from writing $m\omega^{\prime 2} = \zeta$ before? (and using the same trick?)

7. Oct 11, 2012

### JordanGo

Good point! I can also do it that way, I'll try it out.

This problem involves a 2D harmonic oscillator (which the Hamiltonian was for x). To find the degeneracy of the first excited state, can I state that since ωxy=ω, that n = nx + ny.

Then I can write ny=n - x, and set up the following sum:
$\sum(n-n_{x}) = \frac{1}{2}n(n+1)$
where the sum is from nx=0 to n.

So, using the equation to solve for n=1 (first excited state), I get the degeneracy is equal to 1. Is this correct or am I missing something?

8. Oct 11, 2012